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Masters Theses Student Theses and Dissertations

1949

The application of the relaxation method to the solution of The application of the relaxation method to the solution of

problems involving the flow of fluids through porous media problems involving the flow of fluids through porous media

Alexander Antoine Zwierzchowski

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Recommended Citation Recommended Citation Zwierzchowski, Alexander Antoine, "The application of the relaxation method to the solution of problems involving the flow of fluids through porous media" (1949). Masters Theses. 4784. https://scholarsmine.mst.edu/masters_theses/4784

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THE APPLICATION OF THE RELAXATION METHOD TO THE SOLUTION

OF PROBLEMS INVOLVING THE FLOW OF FLUIDS

THROUGH POROUS MEDIA

BY

ALEXANDER A. ZWIEMC HCN/SKI

A

THESIS

submitted to the faculty of the

SCHOOL OF MINES AND METALLURGY OF THE UNIVERSITY OF JaSSOUR!

in partial fulfillment of the work required for the

Degree ot

MASTER OF SCIE~E IN MEX::HANICAL ENGINEERING

Rolla, 140.

1949

-, I··Approved by <'~ C;~

Pi'Ofeasor of Meohanical En in eting/

i

AC KNONLEDGEMENTS

The author is indebted to Professor A. J. Miles for his

timely suggestions and assistance in the solution of the

probl m.

11

iii

PREFACE

The purpose of this investigation is to determine the feasibility

of applying the relaxation method of mathematical analysis to the

solution of problems in rheology. The relaxation method was first

introduced by R. V. Southwell(l) who applied it to the problem of

electrical conducitivity. It Was later used by U. • Emmons(2) to solv

two-dimensional heat conduction problems.

The steady-state flow of fluids through porous media Can be

difficult or impossible by the usual analytical methods even with rather

simple boundary conditions. The relaxation method promises to provide

an easier method of solution and to provide a means of solving som of

the problems not yet solved. It is the object of this thesis to

establish the feasibility of the relaxation method rather than to solve

any new problem. It finds a maximum usefulness in reservoir mechanics

of petroleum engineering.

(1) Southwell, R. V., Proc. Boyal Society, Series A, Vol. 168,PP. 317-350 (1938)

(2) Emmons, H. ., Transactions A.S.M.I., Vol. 63, No.6,pp. 607-61; (Aug. 1943)

CONTENTS

Acknowledgements ... •' .

Preface , .

List of illustrations •••••••.••••..••••••.••.•....•.••.••••..

List of tables ..•..•.••.•••••.••.•...••••••••••••••••••••••••

List of plates ••.•.•••••••.•.••••••••••••••..•.••••••.••••••.

Introduction - A brief discussion of the correspondence

between Heat Conduction and the Flow of

Fluids through Porous Media •••••••••••••••••••

Body - Problem 1 - Solution of simple Radial flow probl m

using the relaxation method of mathematioal analysis ••

Problem 2 - The relaxation method of mathematical

analysis applied to 8 more lengthy Radial flo

problem .

Problem 3 - Solution of Square drainage area prabl

iv

Page

ii

iii

v

vi

vii

1

3

12

by the r laxation method of thematical nalyai us!

a quare network ot flaw ••••••••••••.••••••••••••••••• 21

Problem 4 - Solution of Circular drainage area problem

by the relaxation method of mathematical analysis using

a rectangular network of flow•••••••••••••••••••••..••

Conclusions ••••••.•••.••••••••••••••.•...••••••••.•..••••••••

S \1II1D'lSry- ••••••••••••••••••••••••••••••••••••••••••••••••••••••

Bibliography .. "! .

29

31

39

40

LIST OF ILLUSTRATIONS

y

Fig.

1. Sketch of circular drainage area of problem 1••.••••••••

2. Sketch of circular drainage area of problem 2•••••••••••

3· Sketch of square drainage area of problem 3••••.••••••••

Page

3

12

21

4. Sketch of circular drainage area with rectangular

network of flow of problem 4....................... 295. Enlarged cut view of fig. 4............................. 30

LIST OF TABLES

Table No.

1. Tabulated results and intervening steps in order of

calculation of problem 1 •••••••••••••••••••••••••

2. Tabulated results and intervening steps in order of

calculation of problem 2, method 1••••••••••••••••

3. Tabulated results and intervening steps in order ot

calculation of problem 2, method 2••••••••••••••••

4. Comparison of correct and obtained results in tabular

form problem 2 .•..•••.••••••...•.•..•......•••.•••

,. Tabulat d results and intervening steps in order ot

calculation of problem 3••••••••••••••••••••••••••

6. Tabulated results and intervening steps in order of

calculation ot problem 4••••••••••••••••••••••••••

vi

Page

9

16

18

20

2;

33

vii

LIST OF PLATES

Plate No.

1. Pressure Distribution in Radial flow - problem 1 •••••••••

2. Pressure Distribution in Radiel flow - problem 2,

method 1•........ " .

3. Pressure Distribution in Radial flow - problem 2,

method 2 " " "••• of •••••••••

4. Pressure Distribution in Rectangular flow along west or

east direction from sink - problem 3••••••••••••••••

Page

11

17

19

26

,. Pressure Distribution in Rectangular flow along southeast

direction from sink - problem 3..................... 27

6. Pressure Distribution in Rectangular flow along east-

south-east direction from sink - problem 3.......... 28

7. Pressure Distribution in Rectangular flow due north

from sink - probl m. 4 .•••••••••••....•••.••..•••••••

8. Pressure Distribution in Rectangular flow along northeast

direction from sink - problem 4•••••••••••••••••••••

9. Pressure Distribution in Rectangular tlow of all points

on fig. , - problem 4 •••••••••••••..•••••.••••••••.•

34

35

36

INTRODlX:TION

1

Before illustrating the use of the relaxation method, the

correspondence between heat conduction and the flow of an incompressible

liquid through a porous media will be shown; since the problem of

applying the relaxation method to flow of fluids was attempted when it

was seen that M. W. Emmons(3) has applied it euccessfully to heat

conduction problems.

Heat Conduction

Temperature T

Thermal Conductivity k

Rate of Heat Transfer Q • -ItA fIsothermal Surtace T =Constant

steady state Flow

Pressure P

Permeability !Viscosity u

Velocity Vector V • - ~ ~

Equipressure Surfaoe P - Cons ·t

From above comparison it is seen that Fourier's La is similar to

D'Arcy's Law, the only difference being that temperature is substituted

by pressure.

Also from Laplace's equation, in heat conduction:

= "

and in fluid flow, introducing the velocity potential pt

::.0

(3) Emmons, Ope cit-po iii.

2

Therefore, since the similarity between heat conduction and fluid

flow is clearly shown, it is concluded that the relaxation method will

also apply to problems involving fluid flow.

The main idea in solving steady fluid flow problems by the

relaxation method is that fluid mass in any closed system Can be neither

created nor destroyed. In other words at steady flow conditions the

total quantity, Q, of fluid at any given point at any given instant of

time must add up to zero e.g. 8S much fluid is flowing towards the point

as away from it.

The above idea is the basis for the solution of fluid problems

involving steady state conditions. Essentially the procedure followed

is to assume values of pressure at all points whose preeaure 1s to be

determined, and to calculate Q at any particular point using D'Arcy's

Law. For steady flow Q - 0, so reassign values of pressure until Q

approaches very closely or equals zero.

Four problema involving radial and rectangular flow actually show

the feasibility of using the relaxation method to solve problems of.

similar nature. The radial flow problems can be actually checked by

formula as will be shown subsequently.

PROBLE 1

SOLUTION OF SIMPLE RADIAL FLCYN PROBLEM

USING THE RELAXATION METHOD OF

MATHEMATICAL ANALYSIS

3

As an illustration of the use of the Relaxation Method, consider the

following two-dimensional problem•

.,.,,'

Fig. 1

Given: well bore 100 feetpressure 0 psi

drainage radius 1200 feetpressure 1100 psi

To find pressure at radius of 300', 600', and 900'.

1. Assume straight line variation of pressure with drainag

radius.

2. Flo is radial only.

3. Since drainage area is symmetric 1 in all directions, take

a sample equivalent pipe on any radius 88 shown above.

4

Q·-!A~u ~R

where k = permeability

u • viscosity

A • cross-sectional area of flow

AP = difference in pressuree

AR • difference in radiuses along flow lines

let ! =1 for particular problemu

A N R R· aritbmetic average value of radius

A • 2~R (for unit thickness)27fR p

Q=-AbR

For particular problem

Q::: ,0 I

21T'R P == .2 7T (200) 6 P ... ',:;'8 AP= -.0() , /01> .3-- 6R ;;'00

1 0.2 :t 300' Q.."o I = :2 ". (""5"0) A P'" tf 44 L> P

300

.2 ~ '" 'ooQ<joo' = ;;J.7T' (7So) ~ P == 15.7 ~r>

300

3 0 .. ='1~{)':l.o I ;::. ;1. 7T" (; OS-D) L:."P= :l~.o 6,) P3 (J Q

as shown on Fig. 1

From plate No. 1 assuming straight line variation of pressur with

drainage radius,

P @ 100' =a psi

p @ 3°0' • 200 psi

p @ 600' • 5'00 psi

P @ 900' • 800 psi

For steady flow conditions, total Q • o at any point.

5

Calculate Q with assumed pressures at particular pointe, and it Q1 0

assume other values of pressure until Q=°with correct values of pressure.

a.) At point where r • 300'

Flowing in (+) Q • 9.44 (500-200) • T 2832 unite300 600

Flowing out (-) Q • 6.28 (200-0) • -1256 units100 300

Totel Q • 1" 2832-1256 • + 1576 units

b. ) At point where r l: 600·

Flowing in (+) Q • 15.7 (BOO-Soo) • +4110 units600 900

Flowing out (-) Q • 9.44 (500-200) • -2832 units300 600

Total Q • +4710..2832 .... 1878 units

c.) At point where r • 900'

Flowing in (+) Q = 22.0 (1100-800)='" 6600 units900 1200

Flowing out (-) Q =15.7 (800-500) • -4110 units600 900

Total Q • +6600-4710 • ... 1890 unit

From above it is s en that 8sumed Ta1uee of pressure re inoorrect

for Q , zero for all three points, as Q is not zero for any point.

Therefore using the method of relaxation, that is continually

assuming values ot pressures for a point until equivalent Q values re

qual to or approach zero.

This is shown on table No. 1 where all results are tabulat d in

their order of calculation.

To check results obtained, compare with values of pressure calculated

trom the formula for redial tlo to a sink 88 shown on page 10.

Calculations

Given Po • 0 psi

PI • 200 psi

P2 • 500 psi

P3 = 800 psi

P4 =1200 psi

step 0

Ql • 9.44 (500-200) -6.28 (200-0) =+ 1570

Q2 I: 15.7 (800-500) -9.4-4 (500-200) • +1880

Q3 = 22.0 (1100-800) -15.7 (800-S00) • + 1890

Step I

Change pressure ~3 Bnd make it 8001" 100 • 900 psi

Q3 =22.0 (1100-900) -15.7 (900-500) • -1880

Q2 =15.7 (900-500) -9.44 (500-200)· 34;0

Step II

Change pressure ~ and make it 500+ 200 • 700 psi

Q2 • 15.7 (900-700) -9.44 (700-200) =-1588

'h • 9.44 (700-200) -6.28 (200-0) • 1" 3468

Q3

• 22.0 (1100-900) -15.7 (900-700) • + 1260

Step III

Change pr 8sure OJ. nd make it 200+ 200 =400 psi

Ql • 9.44 (700-400) -6.28 (400-0) aT 318

Q2 • 15.7 (900-700) -9.44 (700-400) -;- 302

6

Step IV

Change pressure C3 and make it 900+ 50 = 950 psi

Q3 • 22.0 (1100-950) -15.7 (950-700) = -635

Q2 : 15.7 (950-700) -9.44 (700-400) -+1087

step V

Change pressure ~2 nd make it 700+80 = 780 psi

Q2 a 15.7 (950-780) -9.44 (780-400) =-824

Ql • 9.44 (780-400) -6.28 (400-0) : +1081

Q3· 22.0 (1100-950' -15.7 (950-780) a+621

Step VI

Change pressure C3 and make it 950-r 20 - 970 psi

Q3 =22.0 (1100-970) -15.7 (970-780)- -133

Q2 • 15.7 (970-780) -9.44 (780-400) • -510

step VII

Change pressure Cl and make it 400 + 70 a 470 pai

Ql z 9.44 (780-470) -6.28 (470-0) • -27

Q2 • 15.7 (970-780) -9.44 (780-470) a + 150

Step VIII

Change pr saure 2 and make it 780-r 8 • 788 psi

Q2 =15.7 ('70-788) -9.44 (788-470) : -51

Q3 =22.0 (1100-910) -15.7 (970-788) : -7

Ql = 9.44 (785-470) -6.28 (470-0) -+ 48

7

Step IX

Change pressure ~1 and make it 470 T 4 • 474 psi

Ql =9.44 (788-474) -6.28 (474-0) • -t 15

Q2 • 15.7 (970-788) -9.44 (188-474) .of 13

Step X

Change pressure @2 and make it 788 -t-l I: 189 psi

Q2 • 15.7 (970-789) -9.44 (789-474) • -12

Ql • 9.44 (189-474) -6.28 (474-0) • ..,. 24

Q3 : 22.0 (1100-970) -15.7 (910-789) =+ 8

Step XI

Change pressure ~1 and make it 4741" 2 • 476 psi

Q1 • 9.44 (789-476) -6.28 (476-0) : -7

QZ • 1,.7 (970-789) -9.44 (789-476) • ..,. 7

8

Tabulated Results

Step QO Po Q1 PI Q2 P2 Q3 P3 Q4 P4-

0 0 +1570 200 +1880 500 +1890 800 lloo1 +3 50 -1880 +1002 +3468 -1588 +200 +12603 +318 +200 +3024- +1087 -63, +50, +1073 -824- +80 +6216 -510 -133 +207 -27 +70 +1,08 -+48 -51 +8 -79 +15 +4- +13

10 -+24 -12 +1 +811 -7 -t2 +7-

Fin 1 Op i 476psi 789psi 970psi noop 1Pressure

Table 1

9

10

It is seen from plate No. 1 that the relaxation method applied to

problem No. 1 resulted in getting values of pressures for point 1, 2,

and 3 very close or identical to thoBe calculated from formula (1) for

radial flow to a well.

where P =pressure @ radius R

Pa z known pressure @ radius Re• 1100 psi

Pw =pressure at face of well bore• 0 psi

Re =drainage radius where pressure is knownI: 1200 f t

Rw = radius of ell• 100 feet

'P =1100- 0 .....

t.... r<)..u. 1:1.00;'00 /00

+0 == 4 o4~ ilIA .1L100

By Relaxation Method

@lR = 300'

@R =600'

@It • 900'

PJ. • 480 psi

P2 .: 784 psi

P3 • 970 psi

PI • 476 p i

P2 = 789 psi

P3 • 970 psi

Comparing actual values ith valu s obtained by usi g the relaxation

method, it is concluded that the relaxation method applied to flow of

fluids will give fairly accurate results tor the problem considered.

It must be noted that had other values of pressure been as umed,

the results would have been slightly altered, as will be shown in

problem No.2.

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PROBLEM 2

THE RELAXATION METHOD OF MATHEMATICAL ANALYSIS

APPLIED TO

A MORE LENGTHY RADIAL FLC1N PROBLEM

s a further illustration of the use of the relaxation method,

consider the followi g two-dimensional radial flow problem.

3.2.0 f'si.

Fig. 2

12

Given w 11 bore 8 inchespressure 100 psi

drainage radius 660 feetpressure 320 pei

Cheek pressure @ points 5',10',25',50',100',200', and 400'from bor hoI •

From D'Arcy'e Lew

where !. 1 (Assumed)u

=2"R (for unit thickness)

For particular problem

13

;l7f(2.3a) a P '" 1.01T",AP

4.~1!>

.. 3.0 '7J.A P

= :lff(J7.!!J. D P z: 3.0 -,r DP;2 ~-

= ,0 I.:100 400

= ,:2""'(150) ~ P "3.0?T ~ P100

:: ;l7f (3 0D) ~ P 5< 3.0 71' c. P.2 DO

= :2 7T' (SJo) L:;) '? : 4. 0" .". .A P:4 '0

As previously Q is ~ for souree (flow into point)

and Q i8 - for sink (flow out of point)

14

From plate No. 2 assuming straight line eriation of pressur ith

drainage radius,

P @1/3' 100 psi

p @5' 101 psi

P @10' 104 psi

P @25' 108 psi

p @50' 116 psi

P @100' 132 psi

P @200' 165 p i

P @400' 232 psi

P @660' 320 psi

Results are shown in ccompenying table No. 2 and plate No.2.

It will be shown that if other than straight line variation of

pressur with drainage radius i assumed, using the relaxation thod

to determine pr saure, will result in very slight (negligibl ) or no

change in final results, 8S is shown on t ble No.4.

From plate No. 3 assuming curv d line variation of pressure with

drainage radius,

P @1/3' 100 psi

p @5' 160 psi

P @10' 200 psi

P C1l25' 220 psi

P 60' 240 psi

P @lOO' 260 psi

P 200' 280 psi

P 00' 300 psi

P @660' 320 psi

Results are shown on table No. 4 and plate o. 3.

Again, the relaxation method applied to th pr c ding r dial flo

problem gives results of pressure very close or identical to those

caleulated from formula (1) for radial flow to a well, as shown below.

15

R"en ­Rw

where P =320 psi

p.. • 100 psi

Re • 660 feet

Rw • 4 inches • 1/3 teet

p = 3.2 0 - 100 ,£n .!i ~ / 0 0 =.::J If "en 3 R + 10 0/} "'~O 'I~

~n-

'iB

Tabulated Results

Step Qo Po QI PI Q2 P2 Q3 P3 Q4 P4 Q5 P, Q6 P6 Q7 P7 Qa Pa0 100 +8 101 -2 104 +17 108 +24 116 +51 132 +102 165 +156 232 3201 +312 -338 +702 +381 -348 i"1l0 -83 +384 -339 +120 1"124 -t377 -336 +120 +21

5 +251 -206 +110 -66 +293 -252 +95 +127 -7 +75 -278 .,.24 -39 +10 +429 -9 -18 +10 +22

10 t-jO -12 +6 +911 -9 +l +312 -11 -7 +7 +913 -tl5 -3 +2 -314 +2 -3 +3 +315 -16 +5 -3 -516 -4 -3 -417 i'3 -3 +1 +318 -2 -3 +1 +

Final lOOps! I73psi 196psi 225psi 246psi 265psi 28,psi 305psi 320psiPressure

Table 2

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Tabulated Results

Step QO Po Q1 PI Q2 P2 Q3 P3 Q4 P4 QS Ps Q6 P6 Q7 P7 Qa Pa

0 100 +60 160 -74 200 +14 220 0 240 0 260 0 280 +21 300 3201 -20 +20 -12 -5 -7 +4 +123 +2 -6 +3 +94 +6 -15 +4 +125 -6 -6 +3 +306 -t'9 -5 +7 0 -3 +2 +18 +5 0 +1 +39 +2 0 +1 +3

10 0 -5 -1311 -9 +3 -3 -712 -1 -2 -313 -5 -1 -1 0

Final 100psi 173psi 197psi 224pBi 244psi 264psi 285psi 305psi 320psiPressure

Table 3

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Tabulated Pressure Values

Correct Metbod Relaxation Method

Firat ssumption First assumptionstraight line curved line

Radius From formula variation variation

5' 171 psi 173 psi 173 psi

10' 198 psi 196 psi 197 psi

25' 225 psi 225 psi 224 psi

50' 245 psi 246 psi 244 psi

100' 265 psi 265 psi 264 psi

200' 285 psi 285 psi 285 psi

400' 305 p i 305 psi 305 psi

Tabl 4

20

PR BLEM 3

SOLUTION OF SQUARE DRAINAGE AREA PROBLEM BY

THE RELAXATIUN METHOD OF MATHEMATICAL ANALYSIS

USING A SQUARE NE'lWORK OF FLOR

Statement of Problem

a.) Use rectangular network instee ot radial as indicated

below.

b.) Given 40 acre spacing or 1320 feet on each side.

21

c.) Pressure @ boundaryPressure @ sink

600 peio pei

/ 3;20 ~e-t

i-JII s/n .t:~ 0nj-,

0t'( (/

m......{.

~

d

c

b

()..

o . I~ (> ~f':"

Fig. 3

3

22

Assumptions

1. .) Since figure is symmetrical, to determine values ot

pressure at all points indicat d (intersection of lines shown

on figure 6), it will be only necessary to determine pressure

at all points on lower right corner bounded by lines a-g and 0-6.

b.) Again, since upper right part of cut figure i symmetrical

to lower left part, pressures are calculated only at half or

given points, .g. point £,2 and e,l are symmetrical to eaoh

other as are £,3 and d,l etc.

2. Consider all rlow in any spacing, concentrated in a pipe shown

as vertical and horizontal lines. For example: segment d,O ­

d,l denotes pipe hieh inelude flow from 10 er half of

pacinI d,O - e,O - a, - d,l and upper half of segment e,O ­

d,O - d,2 - 0,2.

3. ssume values of pressure for giTen points and consider

horizontal and vertical flow, a shown by flow network.

4. Determine Q using assumed pressures, and if Qf °at particular

point, relax by increasing or decreasi g pressures until the

lue of Q equals or pproachea very nearly to zero.

From D' rcy's Lew

where ! • 1 (Assumed)u

=constant (same for all segments)

.:)x. constant (equal lengths of segments)

L) P:II difference in pres ure between t..,o points

where

f =funct ion of

23

24

5

Determination of Q @ point ,3.

step 0

From plate No. 3 sswne pressures as follows:

@ 8,3 P • 390 psi

o e,4 p • 460 psi

d,3 p • 410 psi

o 8,2 p : 330 psi

@ f,3 p = 350 p i

• (460-390) + (410-390) + (330-390) + (350-390)

m -10 units

All result in the order of calculation are shown on table No.5.

Plates No.4, No.5 and No. 6 sho plotting of preeeur vs. distance

in teet from sink.

TABLE ;

Tabulated Results

Qg,l Pg,l Qg,2 Pg,2 Qg,3 Pg,3 Qg,4 Pg,4 Qg,5 Pg,5 Qr,1 Pr,1 Qr,2 Pr ,2 Qr,3 Pr ,3 Qr,4 Pr ,4 Qr,5 Pr,5

417psi519psi466psi393psi 563psi

-20 520+30'­

+100 t

-20" +30+;'"

+15 aC.-t2;.Jr'

-15'" -+10-5"-+5"

+10'"_2S7'

0 11 +3

of 1' '''+3

'S"

-+4 7f

~1'"+2'.

47bpsi 524psi

-10 3;0 + 10 430+130" +100"+2607 +200"-140"+100 -SO' +70-70' -;0"+10''a' 0"-+2;"''' +1;'&'- 35·" +15 +30~~-2;i' -10 +10_lOi? -+ 5"+5,s, +15 J '

+1;1' -25"+10+23"11 -15'"

_94"'1 +8 _8"r~.J, Ofl

: 3""'" -20S'" +5+6.$"1 _11 t7

sr ~

- 6t.' +3 -14"~3" -11 Co'-2 -9

07' _1' '1

4-27S' _15'S' +2

+97' _13 7Co

-1 rJ, +1 -12''''0'" 0'" -3

'-3" ...,1"+1" -1 -2"

+10 270+180.2.+300'-220~+130- 120f'

-10"+5.l.z.

+20"S"-40,21+15_30.21'

-15'"_TrJ

_2$"J

+1£"+4'"-f'6""....2" 1

't-l+3'"+5"+6'"+21'J +1+3"!+2"+1'.t"

343psi

-100 210+200'-280(,+120_20'+10,a.,+40~'

Oa"+lO+12""O~ +3

+2sr

+4 '"'+6"+2,f +1+4'.1-2 rt+2'.r -1

561psi

+40 500-r130"-70$' +;0-10'·+5'&'

+2;'"II'

-I; +10

-~-2+4n+6"+2'1 +1+1"

+60 400+200'-160' +90-110·

+30'_30" +15_1;~1

+5.1'+2;17+3;'Y_;J' +10+2.1'.1

+12$'\ot"? +3

...3""+77.r_11 ' +2

0'77

-H"-;"_1'7 -1

+100 300+270"-290' +140-200f

0'+15'U.+30.21-30t~ T15

ou"+10"+26.s"1_2.r.a +7-+3·'+61'1

+12'"0'2. +3

+2'"+4

7'

0" +1+2'.a

+3'"+2"0"

+140 200+290'-390~+170-250'

-tiO'+-25.&1-35~~ +15- 20.u'

+10'"-+16 ".+23.f'.&.

+3.t"J +5+4.rs'+7'~_1'(, +2+1"+2''''+3"+5'J+1'" +1-2"

+2GO 100-380' +150- 210.a.

+30·_302) +15-1501a

+5.&r-19"" +6-144""_

8r 'l

_12rr +1-10j,(,_14"'" ..n-12"-11f'ri

+1" -3_1'"

Final 270 psiPressure

Qe,2 Pe ,2 Qe,3 Pe ,3 Qe,4 Pe,4 Qe,5 Pe,5 Qd,3 Pd,3 Qd t4 Pd,4 Qd t5 Pd,5 Qc,4 PCt4 Qc,5 PCt5 Qb,5 Pb,5

Final 457psiPressure

569psi 524psi ;76psi 584psi 592psi

+60 560+100 11

0"61'25+20"'"_4'" +6+2$1l

-+4 to

Of' ·n

o 550i"20 11

+70,r-10" +20+1~·+2,''1-1;011'-+10-3-"+3

011

+7'1"-tIl-I'

_ISO +3+1'"+3"-1'" +10·'

;68psi

+80 5001'140'"-60'''' +50,f-20+20s'ao+40·""-8~'+12

+121/</-4'" +4+21"Di'6"<1_2 '0 +2-t2"-2'"

0"_2,J

+10 540+35·'" .+65"-15" 1'20+~If

-t'2;s'Joo_15 tll +10-5U'+5"~

+15 tJ'If/~

-1 +4+2~o

+4$'1+6~"i

+8'"0"" +2

+1"-1',(0"

+1'"0"

551psi

-;0 4900"

+90'''-,(,.

-30 +30_10"+40'''­+;5"-25'·+20_10"

0'-1+12'"-28¥1+10-211/~

-16'"-12"-S'I'-6'"_14,<i -+2_U'.s"_9 1

'

-17" +2-15

'•

-13 101

-10""_ 2"" - 2

-3'"+1" -10'"

+120 410+280 1

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-BoIS'+90-20"+10~'+;0.1-­_10" +1;

+611j1+26"1/+6~' +;

+12"T16'oJ+4(J +3+8 '1

-+12"0'" +3_4"0" -1

_2 fJ

+ 2'Y -1

o 530-t30"+80'1-20'" ....25

0"-n,'"+25'1/-15~.r+10-;:"+2"S'+6'1"+94""-nS1 +2+3"+;,a.+1

7' +1

+31"1_If" +1

o 460+70'

+150''''co J)-..,10 i'50

-25'"+5"

-+1;""+30'"-301

' +1;-10''''

0JS'

+10"+lS~'+28 ~.,

o'if

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+-10 u

+2u+4'" +2+6

'/

+7'~

_1 ' " +2+114"

+3,r-+1 r.t'

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536psi498psi

-10 390+90'

+200"-120 /"- +80

_70'5+20,r

+3;""-25~1 +15-10~o

+5~'

+20JJ

-12'" +8-5~

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+8s '...-4 +3-1~'

+1'"+4 a.4'

+67"

_2" +20"

+31"0+4'"+3"+2,1-+1'"

o 330+260 .,-18011+110-20''''i'10.z,1+40'&'_20"t +15- 4'11

-t2'·+4"+8"07f +2

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NOTE: small numbers to the immediate right of values of ~ denote step number (order or calCUlation).

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PROBLE 4

SOLUTION OF CIRCULAR DRAINAGE AREA PROBLEM BY

THE REJ~TION METHOD OF MATHEMATICAL ANALYSIS

USING A ~TANGULAR NETNORK OF FLOi

29

Finally consider the follo ing problem in olving vertical and

horizontal flow, or a rectangular network appli d to flow fro a

circular boundary to a sink."(I

~r--- 1'00' r--.r---....

// 16"••' "'~ "0• ..../ JilO' f\.

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Fig. 4

Given: sink pressure 0 psi

draina e radius 660 f atpressure 600 psi

To find pressures at all points i dicated (intersections of

vertical and horizontal flow lines).

30

Assumptions

1. Assume str ight line v riation of pressure with dr in ge radius

at all points (shown on plate No.7).

2. a.) Given circular figure is symmetrical, ao consider only one

quarter of circle as shown be10 •

b.) Upper left corner is symmetrical to lower left, so consider

only ower right corner.

3

~o'

,... r-........r---...'~

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c

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f

Fig. 5

3. Flow is Be shown above in arrows e.g. for point c,l to sinkr

direction and path of flo is along se ants c,l - e,O, e,O - b,O,

b,O - a,O and c,l - b,l, b,l - a,l, a,l - a,O.

From D'Arcy's Law

where k • 1 (Assumed)U

=constant (same for all egments)

L!.X = variable

: 100 feet for all segments excepting tho asegments indicated on figure 7.

AP= difference in preae re between two pointswhere Q is considered.

31

32

Sample Calculation

Determination of Q @ point g,2

Step 0

From plates No.7, No.8 snd No.9 8Sum d pressures follow:

g,2 p. 515 psi

C g,l p. 553 psi

f,2 p a 490 psi

boundary p =600 p i

From Q • ~ P , it is noted that Q is inversely proportional to&>x

distance of flow AX • For all points excepting those adjacent to the

boundary, the lengt of path of flow 1s 100 feet. So we may 8sume it

is a constant except for points in question.

hen determining Q at point g,2, we note that vertical distance to

boundary is 28 feet while horizontal distance to boundary is 75 feet.

Since we are not interested in units of Q, but in just its value,

we may neglect entirely ~x for all segments aqua Ing 100 feet in lengt

and use a proportionality constant in case length 1 100 feet.

e.g. at point g,2

• +16 units

s before pressur s ar increased or dec rea ed until all values ot

Q appro ch very nearly or ar qual to roo All results In the order

of calculation are shown on table No.6.

Plates No.7, No.8 and No.9 show plotting of pressure .s. distance

in t at from sink.

TABLE 6

Tabulated Results

Q Pg,O g,O Q Pg,l g,l Q P Q P Q Pg,2 gt2 f,O ftO f,l r,l Qr,2 Pr ,2 Qr,3 Pr,3 Qr,4 Pf',4

+17 545 +16 553 +16 575 +20 454 -+16 464 +16 490 +19 530 +4 583+l07/S' +96/~ -+-86J7 +150,0 +141" +116'''''' +109 / +59'01-33~' +30 +-125

J4 +116.:/-210',{' t90 +231,r +196'" +178 17 +99"

+27,J'~,

+12.z.% +15 -50" -89" "'"80 _8411 +70 +13'P' +40 +5" +10-22J.l +30+3

J' +5 -7.31 -2" 2 _20~o -19 /7 -44" i"22 " +1141

+6"" ,;. + -12.1; +11.11 _29.2.3. -21/1 +6 +1~ +1-2 0_3"D +2 +4 ~. _7Jf -4-18.rS'" -11 Jl -1", +2tl'2. +61.6 +5-¥/l -6 iI' +6 _,~o 0"

+8'1 +90$" _3S"7 +1 'II_2',2 +2 -3..t"1 +3 Or, +3'"

_1,"0 ....2'.1Finel ;82p!i ;a5ps! 592pd ;47psi ;;Opei 560psi 576psi 594pBiPressure

Qe,O Pe,O Qe,l Pe,l Qe,2 Pe,2 Qe,3 Pe,3 Qa,4 Pe,4 Qd,O Pd,O Qd,l Pd,l Qd,2 Pd,2

+2; 363t17, I.

-345/0

+130-95"_;'s'",

+10 .l.a.

+201.2-12J 'I ....8+2~S'

t4S

'i-10.('7-6.s1" -+4_3.J"'-1 "

Final 505psiPressure

+21 375i-171'+301,Q-199" +12;_99',z-19"-4~'

+14"3....22"1(_6J~ +7

O·P+6-t1·+9s'S'

_3$"7 +3+1 t'Y+2'·

S10psi

+20 407+16rf'+285"-115'" +100

-25,1+45.1

+57"-15" +18-8~-4 n-l uo+2'"+3

7J

525psi

+17 455+137<;+23"'~-123'~ +90-68''1-28"_10 3 '_2J

'+4 'II

,oJ-4 ,I' +2-3'1--1

547pai

+20 514+200,3-20'" -+55

O,f+24/1"+6'"+2~t' "'1

570pei

+27 273+187"-413' +150-1137

+17'"+37'""-23"'t' ."'15

+7'"+15",0-2,'.... +10-17'-{_,~7

-1 '13+1 sa+2rl1"8....r

'(. +2+4.)"Y-4" +2_2"

0'10

452psi

+32 287+212'"+362(,-2387 1'150-98'+27"+42,2f-18--' +15-8.t.'"'431

+14'+21"S"

;)7 +6- 3<1'1+1~

+9..rJ_3:$ +3_1:1'+21'7+3 .. ,+5"+1,6 -+1+2 71

+3,a462psi

+26 328+186,s-1"336

7

-224 r' -+140-104'

-4·....i"11~+26.%1-22" +12_4JJ

J'+10"+15.f~-1 tI'1' +4+2'1'+4.r6

+7"+3'" H+470

+67 .....

"13 .H+3"'1

486psi

331psi

Qb,1 Pb,l

+74 129+374'-266.% +1601"94"

_6.&.J -t"2"-10.z.I"+30.t.f

_231 +8+8,r

'It'+18",-20 +51"6-zSl .... 2

$"J­-t-4,r-4 0 +2

0"+2 1

'

260pe1

Qb,O ~ ,0

+-76 911

-524 +150_204.a.-44'

+6a.t.'1

+26.a.Y-6 +8+2"t'

+18:~-24jJ +5+2

+12""1/1-4 +4_2.f"f

+2rl"-3s-t{+5"+9""+ll1 +2

444pei

Qc,2 Pc,2

+32 258+392~-248 t-160+32

Y_28027 +1-~~

+16J1y....

-4 "" +5+4.p-+14",

+2r .a +3-#-8

OS'3 "'2U

+2+4 71

...6'J

+2 ' '1 +1

403psi

Qe,1 Pc,1

...40 204....200."

+:360 '-360: +180-200_507

_2,.l 3

-5~"+lO~"+2;~7-I;:: +10-7+IJ1

37+7iJ."

+12+-16 '13

'It'-4oJ" +5+1,,1+4+6~+8.('1-4 S'.t +3_2oj

rI-1,(",{'

+2+4,1+6,r+2 70 +1+3 11

+41

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+44 182+194

'-446.3 "'160-86"+64"_16,2~ +20

_1.1+7.zT

....27'~9.10

-;IJ. +8+;~r

-tIO-13-6 +4+4+'S'+8'"

s'40;,Z. 1"2

+6S"o+2sY +l.rt,+'4 "+6_21.1 -t"2

010

'+2 7

'

Final 5lopsire89ure

NOTE: small numbers to the immediate right of values of Q denote step number (order of calculation).,

Qd,3 Pd ,3

+22 386+3021'-178' +120

-t2/J

+26'1_63 ' +8+2 oi'l+4 c.J

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CONCLUSIONS

37

From the results obt ined in the solution of probl me involving

steady radial flow to a well, whose solutions are known, it may be seen

that the mathematioal analysis of relaxation can be successfully employed

to the solution of similar problema.

Comparison of results obtained with those obtained by formula show

negligible differences. It must be noted that the length of tim to

solve any given problem will depend primarily on two factors, namely:

1. values of pressure first assumed, and 2. degree and manner in Which

pressures are increased or decreased at particular points. This is

clearly shown in the solution of problem No.2.

From giv n results, it may be seen that the value of Q not always

equals zero, ev n though the particular problem is co.nsidered completed.

An attempt to bring all values of Q down to zero can be undertaken if

more accurate values of pressure are required. otherwi a, depending on

the given data, values of Q slightly great r than or leBa than zero will

result in correct pressur s with one or t most two pounds.

The solutions to problems No. 3 and No. 4 Cannot be checked by

formula, but they may be considered correct ae 1. the curves drawn ar

quite satisfactory and 2. proof that the method will work, from problems

No.1 and No.2.

Problem No. 4 tends to show that no matter what shape of drainage

area one has, one can suocessfully spply the relaxation method of

mathsmatical analysis a surning any desired form of network. The drainage

area in problem No. 4 has a circular boundary, but a rectangular or

rather a square net ork of flo wa assumed and results wers very

sati factory.

38

In conel sion it can be definitely stated that the mathematical

analysi of relaxation will apply to the solution of problems involvin

steady flo of fluids through porOUB media.

StnolARY

39

A very brief review of the similarity between the 8ol~tion of heat­

conduction problema and that of flow of fluids through poro~s media is

made.

The steps folIo ed in the solution of a problem by use of the

relaxation method are shown at the same time solving two radial rl~

problems which are checked for correctness using the formula for radial

flow to a ell.

Two mar problems are solved using the relaxation method, one

involving horizontal and vertical flow in a rectangular boundary, and the

other a circular boundary.

40

BIBLIOGRAPHY

1. Books:

a. Musket, M., The flow of homogeneous fluids through porous edia.

1st ed. N. Y., MCGraw-Hill, 1937. pp. 55-74, 121-140, 149-156.

b. Ing reoll, L. R., Zobel, O. J., and Ingersoll, A. C., Heat

conduction. 1st ed. N. Y., McGraw-Hill, 1948. pp. 1 , 213-216.

2. Periodica s:

a. Christopherson, B. A., and Southwell, R. V., Relax tioD methods

applied to engineering probl ms Ill. Problems involving two

independent veriables. Proceeding of the Royal Society,

series A, Vol. 168, pp. 317-350 (1938).

b. Emmons, H. W., The numerical solution of heat-conduction

problems. Transactions of the A.8.M.E. Vol. 63, No.6,

pp. 607-615 (August 1943).

e. Miles, A. J., and Stephenson, E. A., Pressure distribution in

oil and gas reservoirs by membrane analogy. Amer. rnst. of

Min. and Met. Engrs. Tech. Pub., No. 919 (May, 1938).

VIT

The author was born January 1, 1924, at Istanbul, Turkey. His

high-school education was completed in 1940 at Robert Academy in

Istanbul, Turkey. In 1945 he graduated from Robert College Engineering

School with a B. S. degree in Mechanical Engineering. He entered the

Missouri School of Mines in September, 1947.